from markov inequality:
$$P(X \ge a) \le E(X^q)/a^q$$
$$P(X \ge a) = 1 - P(X \lt a) \le E(X^q)/a^q => P(X < a) \ge 1 - E(X^q)/a^q$$ from
https://en.wikipedia.org/wiki/Generalized_gamma_distribution
https://en.wikipedia.org/wiki/Gamma_distribution#General
$$E(X^q) = a \Gamma((d+1)/p)/\Gamma(d/p) = |where:d=n/q, a=(1/3)^q, p=1/q|=(1/3)^q\Gamma(n + q)/\Gamma(n)$$ now: $$E(X^q)/a^q = (1/3)^q\Gamma(n + q)/\Gamma(n)*2^q/n^q =\\ = (\frac{2}{3})^q \frac{\Gamma(n+q)}{\Gamma(n)n^q}\stackrel{for\ q<n/4\ and\ big\ n}<(\frac{2}{3})^q (\frac{5}{4}n)^q/n^q = (\frac{10}{12})^q \stackrel{q -> \infty}\rightarrow 0$$ $$P(X < a) \ge 1 - E(X^q)/a^q \ge 1 -(\frac{10}{12})^q$$ and two easy steps...
There are some minor mistakes but whole idea should be correct